To solve this problem, let's break it down step-by-step:
1. Height of the Pyramid:
The length of the base edge of the pyramid is given as [tex]\( x \)[/tex].
The height of the pyramid is 3 times this base edge length. Therefore, the height of the pyramid is:
[tex]\[
3x
\][/tex]
2. Area of an Equilateral Triangle:
The area of an equilateral triangle with side length [tex]\( x \)[/tex] is given by:
[tex]\[
\frac{x^2 \sqrt{3}}{4}
\][/tex]
3. Area of the Hexagon Base:
A regular hexagon can be divided into 6 equilateral triangles. Therefore, the total area of the hexagon base is 6 times the area of one equilateral triangle:
[tex]\[
6 \times \frac{x^2 \sqrt{3}}{4} = \frac{6x^2 \sqrt{3}}{4} = \frac{3x^2 \sqrt{3}}{2}
\][/tex]
4. Volume of the Pyramid:
The volume of a pyramid is given by:
[tex]\[
\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\][/tex]
Substituting the known values:
[tex]\[
\text{Volume} = \frac{1}{3} \times \left( \frac{3x^2 \sqrt{3}}{2} \right) \times 3x
\][/tex]
Simplify the expression:
[tex]\[
\text{Volume} = \frac{1}{3} \times \frac{3x^2 \sqrt{3}}{2} \times 3x = \frac{3 \times 3 x^3 \sqrt{3}}{2 \times 3} = \frac{9x^3 \sqrt{3}}{6} = \frac{3x^3 \sqrt{3}}{2}
\][/tex]
So, the detailed answers are:
- The height of the pyramid is [tex]\( 3x \)[/tex].
- The area of an equilateral triangle with side length [tex]\( x \)[/tex] is [tex]\( \frac{x^2 \sqrt{3}}{4} \)[/tex] square units.
- The area of the hexagon base is [tex]\( \frac{3x^2 \sqrt{3}}{2} \)[/tex] square units.
- The volume of the pyramid is [tex]\( \frac{3x^3 \sqrt{3}}{2} \)[/tex] cubic units.
